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djyliett [7]
3 years ago
13

Martha tries to exercise at least 30 minutes eac day write an inequality to represent the number of minutes Martha exercises eac

h day
Mathematics
1 answer:
Sergio [31]3 years ago
4 0
Let “t” represent the number of minutes Martha exercises each day. If she needs to exercise *at least* 30 minutes, t has to be equal to or greater than 30, so our inequality would be

t ≥ 30
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Write g(x) = 4x2 + 88x in vertex form. The function written in vertex form is g(x) = (x +11)2 + .
romanna [79]

Answer:

y = 4(x + 11)² - 484

Step-by-step explanation:

y = 4x² + 88x

factor the expression

y = 4(x² + 22x)

complete the square

y + ? = 4(x² + 22x + ?)

y + ? = 4(x² + 22x + 121)

add 4 • 121 to the left side

y + 4 • 121 = 4(x² + 22x + 121)

multiply

y + 484 = 4(x² + 22x + 121)

y + 484 = 4(x + 11)²

subtract both sides by 484

y = 4(x + 11)² - 484

6 0
3 years ago
Read 2 more answers
Jose owes his brother $50. He promises to pay his brother back $5 each week until his debt is paid off. Represent the relationsh
Maru [420]

Answer:

it will take 10 weeks for him to pay his brother completly

Step-by-step explanation:

3 0
3 years ago
What is the measure of
Kamila [148]

Answer:

64.4

Step-by-step explanation:

In one of the trig formulas, it states    sin <a/A = sin<b/B = sin<c/C

So we have in this case:

sin(70)/12 = sin x/14           Note: x is just a variable for the angle <ABC

14(sin70)/12 = sin x

sin^-1(14(sin70)/12) = x

x=64.4

Your welcome, and comment if you have any questions! :D

6 0
2 years ago
PLEASE HELP GIVONG BRAINLIEST PLSSSSSSSHDHFHFHFJFJJFJRJRHR AHHHHHH
34kurt

Hi there hopefully this helps!

-----------------------------------------------------------------------------------------------------

Answer: C.

Explanation:

A New Leash On Life's Graph has: 3, 11, 21, 13, and 4.

The mean: 10.4

No Ruff Stuff's graph has: 1, 9, 16, 16, 8, 1, and 1.

The mean: 7.43.

3 0
3 years ago
Use L’Hospital’s Rule to evaluate the following limit.
Serga [27]

Answer:

3

Step-by-step explanation:

lim(t→∞) [t ln(1 + 3/t) ]

If we evaluate the limit, we get:

∞ ln(1 + 3/∞)

∞ ln(1 + 0)

∞ 0

This is undetermined.  To apply L'Hopital's rule, we need to rewrite this so the limit evaluates to ∞/∞ or 0/0.

lim(t→∞) [t ln(1 + 3/t) ]

lim(t→∞) [ln(1 + 3/t) / (1/t)]

This evaluates to 0/0.  We can simplify a little with u substitution:

lim(u→0) [ln(1 + 3u) / u]

Applying L'Hopital's rule:

lim(u→0) [1/(1 + 3u) × 3 / 1]

lim(u→0) [3 / (1 + 3u)]

3 / (1 + 0)

3

4 0
3 years ago
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